In: Statistics and Probability
Does living together cause divorce? Couples were given a standardized psychological assessment that indicates the likelihood of divorce (higher scores indicate a higher likelihood of divorce) and asked them how many months they've lived together. The data is reported below:
Likelihood of divorce score | Months lived together |
87 | 54 |
65 | 36 |
45 | 20 |
32 | 5 |
05 | 0 |
Please complete all steps of hypothesis testing, create a scatterplot, and write a short explanation for what you think may be the underlying factor in this correlation (hint: be careful not to associate causal factors).
We now have the following data:
Divorce | Lived |
87 | 54 |
65 | 36 |
45 | 20 |
32 | 5 |
5 | 0 |
As such we get the higher scores indicates higher likelihood of divorce. In order to assess this, we shall calculate the correlation coefficient and then test it using testing of Hypothesis.
We calculate the correlation coefficient using the formula:
To calculate this quantity, we form the following table:
87 | 54 | 7569 | 2916 | 4698 | |
65 | 36 | 4225 | 1296 | 2340 | |
45 | 20 | 2025 | 400 | 900 | |
32 | 5 | 1024 | 25 | 160 | |
5 | 0 | 25 | 0 | 0 | |
Total | 234 | 115 | 14868 | 4637 | 8098 |
We shall substitute the values into the formula:
We can see that the correlation coefficient is 0.9723. This indicates the linear relationship between likelihood of divorce and length of living together.
Null hypothesis: There is no correlation between likelihood of divorce score and months living together.
Alternate Hypothesis: There is a correlation.
Level of significance
Test statistic : follows a t distribution with n-2 df.
The critical value of t at 3 df=3.1825 or the p-value is 0.0055. This can be found by T.INV.2T(0.05,3) function in EXCEL.
Since the calculated value of t>the critical value, we reject the null hypothesis and hence the correlation is present between likelihood of divorce score and months living together